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Theorem sspsstri 3907
Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)
Assertion
Ref Expression
sspsstri ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))

Proof of Theorem sspsstri
StepHypRef Expression
1 or32 940 . 2 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
2 sspss 3904 . . . 4 (𝐴𝐵 ↔ (𝐴𝐵𝐴 = 𝐵))
3 sspss 3904 . . . . 5 (𝐵𝐴 ↔ (𝐵𝐴𝐵 = 𝐴))
4 eqcom 2813 . . . . . 6 (𝐵 = 𝐴𝐴 = 𝐵)
54orbi2i 927 . . . . 5 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
63, 5bitri 266 . . . 4 (𝐵𝐴 ↔ (𝐵𝐴𝐴 = 𝐵))
72, 6orbi12i 929 . . 3 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
8 orordir 944 . . 3 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
97, 8bitr4i 269 . 2 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵))
10 df-3or 1101 . 2 ((𝐴𝐵𝐴 = 𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
111, 9, 103bitr4i 294 1 ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 197  wo 865  w3o 1099   = wceq 1637  wss 3769  wpss 3770
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-3or 1101  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-ne 2979  df-in 3776  df-ss 3783  df-pss 3785
This theorem is referenced by:  ordtri3or  5968  sorpss  7172  sorpssi  7173  funpsstri  31985
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